Mathematics Asked by Ali Shadhar on November 15, 2020

I am not sure if there exists a closed form for

$$I=int_0^{pi/2}frac{x^2 text{Li}_2(sin^2x)}{sin x}dx$$

which seems non-trivial.

I used the reflection and landen’s identity, didn’t help much.

In case you are curious how I came up with this integral:

From here we have

$$arcsin^3(x)=6sum_{k=1}^inftyleft[sum_{m=0}^{k-1}frac{1}{(2m-1)^2}right]frac{{2kchoose k}}{4^k}frac{x^{2k+1}}{2k+1},quad |x|<1$$

differentiate both sides with respect to $x$ we get

$$frac{arcsin^2(x)}{sqrt{1-x^2}}=2sum_{k=1}^inftyleft[sum_{m=0}^{k-1}frac{1}{(2m-1)^2}right]frac{{2kchoose k}}{4^k}x^{2k}$$

use $sum_{m=0}^{k-1}frac{1}{(2m-1)^2}=H_{2k}^{(2)}-frac14H_k^{(2)}$ and replace $x$ by $sqrt{x}$ we get the form

$$frac{arcsin^2(sqrt{x})}{sqrt{1-x}}=2sum_{k=1}^inftyleft[H_{2k}^{(2)}-frac14H_k^{(2)}right]frac{{2kchoose k}}{4^k}x^{k}$$

Divide both sides by $x$ then $int_0^y$ we have

$$int_0^yfrac{arcsin^2(sqrt{x})}{xsqrt{1-x}}dx=2sum_{k=1}^inftyleft[H_{2k}^{(2)}-frac14H_k^{(2)}right]frac{{2kchoose k}}{4^k}frac{y^{k}}{k}$$

Next, multiply both sides by $-frac{ln(1-y)}{y}$ then $int_0^1$ and use $-int_0^1 y^{k-1}ln(1-y)dy=frac{H_k}{k}$

$$2sum_{k=1}^inftyleft[H_{2k}^{(2)}-frac14H_k^{(2)}right]frac{{2kchoose k}}{4^k}frac{H_k}{k^2}=-int_0^1int_0^yfrac{arcsin^2(sqrt{x})ln(1-y)}{xysqrt{1-x}}dxdy$$

$$=int_0^1frac{arcsin^2(sqrt{x})}{xsqrt{1-x}}left(-int_x^1frac{ln(1-y)}{y}dyright)dx$$

$$=int_0^1frac{arcsin^2(sqrt{x})}{xsqrt{1-x}}left(zeta(2)-text{Li}_2(x)right)dx$$

$$overset{sqrt{x}=sintheta}{=}2int_0^{pi/2}frac{x^2}{sin x}(zeta(2)-text{Li}_2(sin^2x))dx$$

So we have

$$sum_{k=1}^inftyleft[H_{2k}^{(2)}-frac14H_k^{(2)}right]frac{{2kchoose k}}{4^k}frac{H_k}{k^2}=zeta(2)int_0^{pi/2}frac{x^2}{sin x}dx-int_0^{pi/2}frac{x^2 text{Li}_2(sin^2x)}{sin x}dx$$

The first integral can be calculated by applying integration by parts then using the fourier series of $ln(tanfrac x2)$.

Another question is, clearly the two sums on the LHS are convergent as the denominator blows to infinity much faster than the numerator. But does there exist a closed form for each?

All methods are welcome. Thank you

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